[PDF] The complex inverse trigonometric and hyperbolic functions

Indeed, if we write

Starting from

Arccotz=12iLn?z+iz-i?

Arccot(z) = Arctan?1

Settingz=x, wherexis real,

Arccotx=1

2Arg?x+ix-i?

2 It is instructive to derive another relation between the arctangent and arccotangent functions. First, we first recall the property of the multi-valued complex logarithm, ln(z1z2) = ln(z1) + ln(z2),(4) as a set equality. It is convenient to define a new variable, v=i-z i+z,=? -1v=z+iz-i.(5)

It follows that:

arctanz+ arccotz=1 2i? lnv+ ln? -1v?? =12iln?-vv? =12iln(-1). Since ln(-1) =i(π+ 2πn) forn= 0,±1,±2..., we conclude that arctanz+ arccotz=12π+πn,forn= 0,±1,±2,... To obtain the corresponding relation between the principalvalues, we compute

Arctanz+ Arccotz=1

2i?

Lnv+ Ln?

-1v?? 1 2i?

Ln|v|+ Ln?1|v|?

+iArgv+iArg? -1v?? 1 2?

Argv+ Arg?

-1v?? .(6) It is straightforward to check that for any non-zero complexnumberv,

Argv+ Arg?

-1 v?

π,for Imv≥0,

-π ,for Imv <0.(7)

Using eq. (5), we can evaluate Imvby computing

i-z i+z=(i-z)(-i- z) (i+z)(-i+z)=1- |z|2+ 2iRez|z|2+ 1 + 2Imz.

Writing|z|2= (Rez)2+ (Imz)2in the denominator,

i-z i+z=1- |z|2+ 2iRez(Rez)2+ (Imz+ 1)2. 3

Hence,

Imv≡Im?i-z

i+z? =2Rez(Rez)2+ (Imz+ 1)2.

We conclude that

Imv≥0 =?Rez≥0,Imv <0 =?Rez <0.

Therefore, eqs. (6) and (7) yield:

Arctanz+ Arccotz=???1

2π ,for Rez≥0,

1

2π ,for Rez <0.

(8) ?CAUTION!! Many books do not employ the definition for the principal value of the inverse cotangent function given above. Instead, they choose to define:

Arccotz=π

2-Arctanz .(9)

Note that with this definition, the principal interval of thearccotangent function for real valuesxis

0≤Arccotx < π ,

instead of the interval quoted in eq. (3). One advantage of this latter definition is that the real-valued inverse cotangent function, Arccotxis continuous atx= 0, in contrast to eq. (8) which exhibits a discontinuity atx= 0.?On the other hand, one drawback of the definition of eq. (9) is that eq. (2) is no longer respected. Additional disadvantages are discussed in refs. 2 and 3, where you can find a detailed set of arguments (from the computer science community) in supportof the conventions adopted by Abramowitz and Stegun (ref. 1) and employed in these notes.

2. The inverse trigonometric functions: arcsin and arccos

The arcsine function is the solution to the equation: z= sinw=eiw-e-iw 2i. ?In our conventions, the real inverse tangent function, Arctanx, is a continuous single-valued function that varies smoothly from-1

2πto +12πasxvaries from-∞to +∞. In contrast, Arccotx

varies from 0 to-1

2πasxvaries from-∞to zero. Atx= 0, Arccotxjumps discontinuously

up to 1

2π. Finally, Arccotxvaries from12πto 0 asxvaries from zero to +∞. Following eq. (8),

Arccot 0 = +π/2, although Arccotx→ -1

2πasx→0-(which means asxapproaches zero from

the negative side of the origin). 4

Lettingv≡eiw, we solve the equation

v-1 v= 2iz . Multiplying byv, one obtains a quadratic equation forv, v

2-2izv-1 = 0.(10)

The solution to eq. (10) is:

v=iz+ (1-z2)1/2.(11) Sincezis a complex variable, (1-z2)1/2is the complex square-root function. This is a multi-valued function with two possible values that differby an overall minus sign. Hence, we do not explicitly write out the±sign in eq. (11). To avoid ambiguity, we shall write v=iz+ (1-z2)1/2=iz+e1

2ln(1-z2)=iz+e12[Ln|1-z2|+iarg(1-z2)]

=iz+|1-z2|1/2ei

2arg(1-z2).

By definition,v≡eiw, from which it follows that w=1 ilnv=1iln? iz+|1-z2|1/2ei

2arg(1-z2)?

The solution toz= sinwisw= arcsinz. Hence,

arcsinz=1iln? iz+|1-z2|1/2ei

2arg(1-z2)?

The principal value of the arcsine function is obtained by employing the principal value of the logarithm and the principle value of the square-root function (which corresponds to employing the principal value of the argument). Thus, we define

Arcsinz=1iLn?

iz+|1-z2|1/2ei

2Arg(1-z2)?

We now examine the principal value of the arcsine for real-valued arguments. Settingquotesdbs_dbs2.pdfusesText_2